Friction solved problems

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According to the second law of motion we have	F = m a
In the gravitational system the force acting on a mass is	W = m g [kgm m / s² ]
A force of 1 Newton is equal to	N = 1 [kgm m / s² ]
Then if the weight is given it equals	W [kgf ] = 9.8 N

There are two friction coefficients the static μ_s and the the dynamic or kinetic μ_k

The static friction coefficient value is greater then the kinetic coefficient value ( μ_s > μ_k ), it means that we have to employ greater force in order to start the movement of the mass, but once the mass is moving we have to employ less force to keep it moving.

In many problems we assume that both coefficients are the same, if not we have to relate to the proper coefficient e.g. if the bodies are moving we have to take the kinetic friction coefficient.

In many questions we have to decide the direction of the movment, it is important because the direction of the friction force is always opposite to the movment direction.

If we look at the figure at left the direction of the movment is to the right as the direction of a₁ the equilibrium equation will be:

ΣF_x = F − m g sin θ − f = m a

and the acceleration is:

If the movment is downward as in the second figure at left then the equilibrium equation will be:

ΣF_x = m g sin θ − F − f = m a

and the acceleration is:

We can clearly see the difference of the signs and the values of the accelerations in both direction of the chose directions.

The conclusion frome this calculation is that in friction problems a negative acceleration result doesn't mean that the acceleration is in the opposite direction but rather we have to change the direction of the acceleration and solve again, if this time we got negative value then the mass will not move due to the friction force, for more information see example 7, example 38.

If mass M is moving a distance of x meter then this distance is divided by each cable that holds mass m and mass m is moving a distance of x/2 meter.

The distance traveled is given by: x = v₀ t + a t² / 2

At rest v₀ = 0 and the acceleration is: a = 2 x / t²

time t of both masses M and m is the same so we can see that the acceleration is related to the distance, if the distance is half also the acceleration will be half so: a₁ = 2 a₂

For system rest we need that the acceleration will be 0. From the equation v_t = v₀ + a t we get the result v_t = v₀

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A body whose mass is m is resting on an inclined surface with an angle of θ and a friction coefficient of μ, find the acceleration of the mass m as a function of μ and angle θ.

From equilibrium in the x and y direction we get:

ΣF_x = m g sinθ − N μ = m a	(1)
ΣF_y = N − m g cosθ = 0	(2)

From eq. (2)	N = m g cosθ
Substitute N into eq. (1)	m g sinθ − m g μ cosθ = m a
And the acceleration is:	a = g( sin θ − μ cos θ)

When the system is at rest a = 0 then:

We can see that the friction coefficient for rest condition must be equal or greater then the tangent of the slope μ > tan θ.

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A body whose mass is m is resting on a horizontal surface whose friction coefficient is μ, a force F is implied on the mass at an angle of θ degree. Find the acceleration of mass m and the maximum force that the mass m will still be at rest.

From the forces diagram we can derive the equilibrium equations:

ΣF_x = F cos θ − Nμ = ma	(1)
ΣF_y = N + F sin θ − mg = 0	(2)

From eq. (2)	N = mg − F sin θ
Substitute N into eq. (1)	F cos θ − mgμ + Fμ sin θ = ma

Solving for a we get:

(3)

For the maximum force F that can be applied so that the mass m will still be at rest we have:

f_max = N μ_s = (m g − F sinθ) μ_s	(4)
	(5)

A force of 30 N is applied to a mass of 3 kg for 5 sec at an angle of 30 degree, if the static and kinetic frictions are 0.5 and 0.2 respectively. Find the acceleration and the time needed for mass m to stop after force F is eliminated.

First check if the mass will move by the condition: f_max < F cosθ.

Because 7.2 < 25.98 the mass will move and the acceleration is according to equation (3).

The velocity of the mass after 5 sec is found from the acceleration aquation:

v_t = v₀ + a t = 0 + 7.7 * 5 = 38.5 m/s

The new free body diagram after eliminating force F is at left.

According to the second law of motion F = m a (F now is only the friction force) and f = − m a (the minus sign is because of the opposite directions of a and f).

a = − f / m = − m g μ_k / m = −g μ_k = − 9.8 * 0.2 = − 1.96 m/s²

And the time untill mass m will stop is according to the equation: v_t = v₀ + a t where v_t = 0 and v₀ = 38.5 m/s

t = − v₀ / a = − 38.5 / − 1.96 = 19.64 sec

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A body whose mass is m is resting on a slope surface whose friction coefficient is μ and with an angle of θ, a force F is implied on the mass at an angle of α degree. Finf the acceleration of mass m and the maximum force that the mass m will still be in rest.

We will choose a positive acceleration upward and define δ = α + θ.

From the forces diagram we can derive the equilibrium equations:

ΣF_x = F cos δ − mg sin θ − Nμ = ma	(1)
ΣF_y = N − mg cos θ − F sinδ = 0	(2)

From eq. (2) we have:

N = mg cos θ + F sin(α + θ)

If the result of a is negative then the acceleration is in the opposite direction.

The maximum force for rest condition (when a = 0) is:

Check the case that the direction of the force F is upward as shown below.

The friction force is: f = Nμ = (F sinδ − m g cosθ) μ (angle δ = α − θ)

From the forces diagram we can derive the equilibrium equations in the slope direction, we assume that the motion is upward if the result of a is negative then we have to change the directiom of the motion.

ΣF_x = F cosδ − f − mg sin θ = m a

(3)

and the acceleration is:
If a = 0 then:

Notice that we got the same answer as before but with some variations in the signs.

Check the case that the direction of the force F is downward as shown below.

The friction force is: f = Nμ = ( m g cosθ + F sinδ) μ (angle δ = α − θ)

From the forces diagram we can derive the equilibrium equations in the slope direction:

ΣF_x = F cosδ − f + mg sin θ = m a

(4)

and the acceleration is:

(5)

Find the force F that should be applied in order to bring the weight of 100 kg to a velocity of 30 m/s in 20 m neglect the friction.

From the acceleration equations we can find the acceleration: v_t² = v₀² + 2 a d

After substituting v₀ = 0 we get a = v_t² / (2 d) = 900 / (2 * 20) = 22.5 m / s²

From eq. (5) we get:

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Two masses are located on each other, mass m is tied to the wall by a rope, a force F is exerted on mass M , find the ecceleration of mass M if the friction coefficients between the two masses is μ₁ and μ₂ between mass M and the lower surface.

Friction force f₁ is:	f₁ = mgμ₁
Friction force f₂ is:	f₂ = (M + m)gμ₂

Mass M is effected by both frictions f₁ and f₂ and is:

f_M = f₁ + f₂ = mgμ₁ + (M + m)gμ₂

From the forces diagram we can derive the equilibrium equations in the x direction on both masses:

On mass m:	ΣF_x = T₁ − mgμ₁ = 0	(1)
On mass M:	ΣF_x = F − f₁ − f₂ = Ma	(2)

Maximum force F that can be applied to mass M without sliding is:

F <= mgμ₁ + (M + m)gμ₂

From eq (1) T₁ can be found:

T₁ = mgμ₁

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Mass m₁ is resting on mass m₂ and has a friction coefficients of μ_1S μ_1k the friction coefficients of m₂ and the lower surface are μ_2S μ_2k . Find the biggest force F that can be applied to m₂ without causing mass m₁ to slip.

m₁		kg
m₂		kg
μ_1S
μ_1k
μ_2S
μ_2k
F		N
a₁		m/s₂
a₂		m/s₂
f₁		N
f₂		N

Force needed to start moving the masses is: F = (m₁ + m₂) g μ_2S

Mass m₁ is attached to m₂ only by the maximum friction force:
f₁ = m₁ g μ_1S this force is causing mass m₁ to accelerate to its maximum value before slipping:

ΣF_x on m₁

m₁ g μ_S1 = m₁ a₁

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a₁ = g μ_1S

(1)

From the free body diagram of mass m₂ we have

ΣF_x on m₂	F − f₁ − f₂ = m₂ a₂
	F − m₁ g μ_1S − (m₁ + m₂) g μ_2k = m₂ a₂	(2)

Solving for the value of a₂ from eq. (2) we get:

(3)

In order to prevent from mass m₁ to slip both accelerations a₁ and a₂ should be the same a₁ = a₂ and the maximum force F is:

F = (m₁ + m₂)(a₁ + g μ_2k)

(4)

If the applied force F is greater then the value of F in eq. (4) then the mass m₁ will slide on top of mass m₂ and the friction force f₁ will be the dynamic force which is less then the static force because μ_s > μ_d so the forces on mass m₂ in this case is:

F − m₁gμ_1k − (m₁ + m₂)gμ_2k = m₂ a₂

Q. Mass of 3 kg is mounted on top of a 7 kg mass, the friction coefficients of the upper mass are μ_1S = 0.8 and μ_1k = 0.1 and the lower mass friction coeficients are μ_2S = 0.2 and μ_2k = 0.1 . Find the maximun force F in case that mass m₁ will not slide.

S. First find the maximum acceleration acting on m₁ with no sliping according to eq. (1):

a₁ = g μ_1S = 9.8 * 0.8 = 7.84 m/s²

This is also the maximum acceleration of mass m₂ according to eq. (4) with the assumption that a₁ = a₂

F = (m₁ + m₂)(a₁ + g μ_2k) = 88.2 N

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Find the acceleration of the system of masses neglecting the mass of the string and the inertia of the pulley.

The kinetic friction force is equal to:

f = μ_k N = μ_k M g

From the free body diagram on mass m and M and assuming that the acceleration is downward we get:

On mass M:	ΣF_x = T − M g μ_k = M a	(1)
On mass m:	ΣF_y = m g − T = m a	(2)

Eliminating T from both equations gives:

m g − M g μ_k + M a = −m a
	(3)

Mass M will accelerate when	m > μ_s M
System of the masses are at rest when	m <= μ_s M (then a = 0)

Find the friction coefficient and the tension in the rope if M = 5 kg and m = 1 kg the masses are moving at a constant speed.

Because the masses are moving at a constant speed it is necessary that a = 0. from eq. (3) we have m − M μ_k = 0

The tension in the rope can be found by solving eq. (1) and (2) :

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Find the acceleration of the system of masses, the masses of the rope and the pulleys are negligible, assume first that the acceleration of M is upward and second time that the acceleration is downward.

The friction force is equal to:

f = μ N = μ M g cosθ

From the forces diagram on mass m and M and assuming that the acceleration is upward (Figure 1) we get:

On mass M:	ΣF_x = T − M g sinθ − f = M a	(1)
On mass m:	ΣF_y = m g − T = m a	(2)

(3)

Solving the case with the acceleration downward (Figure 2).

On mass M:	ΣF_x = M g sinθ − T − f = M a	(4)
On mass m:	ΣF_y = T − m g = m a	(5)

(6)

Check the case when M = 2 kg m = 1.2 kg angle θ = 30 degree and the friction coefficient is 0.15

Because both accelerations in both directions are negative motion is not possible in this conditions of masses, friction and slope angle (see note 2).

Determine the acceleration and the motion direction if M = 4 kg
m = 2 kg angle θ = 45 degree and the friction coefficient is 0.1

First check the case of upward acceleration (Figure 1).

Because a is negative and the present of friction we have to make the calculation again assumming this time that the acceleration is downward see the case of (Figure 2).

Notice that we got a different value for the acceleration, the conclution is that mass M is slidding downward.

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Figure 1

Figure 2

M		kg
m		kg
θ		deg
μ
a		m/s₂
T		N
f		N

The friction force is equal to:

f = μ N = μ M g cosθ

From the forces diagram on mass m and M and assuming that the acceleration is upward (Figure 1) we get:

On mass M:	ΣF_x = T − M g sinθ − f = M a	(1)
On mass m:	ΣF_y = m g − T = m a	(2)

(3)

Solving the case with the acceleration downward (Figure 2).

On mass M:	ΣF_x = M g sinθ − T − f = M a	(4)
On mass m:	ΣF_y = T − m g = m a	(5)

(6)

Check the case when M = 2 kg m = 1.2 kg angle θ = 30 degree and the friction coefficient is 0.15

Because both accelerations in both directions are negative motion is not possible in this conditions of masses, friction and slope angle (see note 2).

Determine the acceleration and the motion direction if M = 4 kg
m = 2 kg angle θ = 45 degree and the friction coefficient is 0.1

First check the case of upward acceleration (Figure 1).

Because a is negative and the present of friction we have to make the calculation again assumming this time that the acceleration is downward see the case of (Figure 2).

Notice that we got a different value for the acceleration, the conclution is that mass M is slidding downward.

Note 1: from eq. (3) and (6) and comparing the acceleration to zero we can derive the range of mass m

that the system will stay at rest. M(sinθ − μ cosθ) < m < M(sinθ + μ cosθ)

Note 2: to determine the direction of the motion we can solve the equilibrium equation by eliminating the friction, once we have the direction of the motion we can set the correct direction of the friction force (always opposite to the motion direction) and solve the equilibrium equations. We have to remember that if we found motion in any direction without friction, the friction force can reduce the motion acceleration or even stope the motion.

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Find the acceleration of the system of masses neglecting the mass of the string and the inertia of the pulley.

From the forces diagram on mass m and M and assuming that the acceleration is downward hance negative we get:

On mass M:	ΣF_x = T − Mgμ₁ = −Ma	(1)
On mass m:	ΣF_y = mg sinθ + mgμ₂ cosθ − T = −ma	(2)

Substitute the value of T from eq. 1 to the second eq.

System is at rest when:

Mμ₁ − m(sinθ + μ₂ cosθ) = 0

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Free body diagram when acceleration is to the right

Find the acceleration of the system of masses neglecting the mass of the rope and the inertia of the pulley, the frictions coefficients and the slopes are given, also find the tension in the rope.

To determine the possible direction of the motion we will first solve the forces equations by neglecting the frictions, the results are the conditions:

m sinβ > M sinα	motion can be to the right	(a)
m sinβ < M sinα	motion can be to the left	(b)
m sinβ = M sinα	motion not possible	(c)

Those conditions alone are not enough to verify if the system will move, in order to make the system to slipe the masses have to overcome the friction forces, so the conditions for motions are.

m ( sinβ − μ_2s cosβ ) > M ( sinα + μ_1s cosα )	motion is to the right	(d)
M ( sinα − μ_1s cosα ) > m ( sinβ − μ_2s cosβ )	motion is to the left	(e)

From the forces diagram and assuming that the acceleration is to the right we get:

On mass M:	ΣF_x = T − M g sinα − M g cosα μ_1k = M a	(1)
On mass m:	ΣF_x = m g sinβ − T − m g cosβ μ_2k = m a	(2)

Acceleration when motion is to the right

System is at rest when:

m(sinβ − μ_2scosβ) − M(sinα + μ_1s cosα) = 0

In order to get positive tension in the rope we need that at least one of the equations will be true:

μ_1s < tanα

μ_2s < tanβ

Free body diagram when acceleration is to the left

Given two masses M = 40 kg and m = 20 kg, connected with an ideal pulley, the masses are located on two surfaces whose friction coefficients are μ_1s = μ_1k = 0.15 and μ_2s = μ_2k = 0.25, the slopes has angles of α = 30 deg and β = 53 deg. Find the direction and the value of the acceleration and the tension in the rope which connects both masses.

M		kg
α		deg
m		kg
β		deg
μ_1s
μ_1k
μ_2s
μ_2k
a		m/s²
T		N

Checks the direction of motion in the case that
the friction is eliminated from the system.

Each mass move

Checks if motion occurs for each mass separately in the case that the rope connecting both masses is eliminated and only friction forces are present μ < tanα

* Green M is to mark that M will slide.
* Red m is to mark that m will not slide.

Describes if the system of masses will slide to the left or to the right or it will not slide at all.

Notes:

1.	When system moves to the right or left the calculation of the tension is straightforward according to the equations found before.
2.	When the system has no motion the maximum tension can be from 0 to the minimum static friction force calculated separately for each mass (see note 3).
3.	In the case of no motion we have to analyze the conditions of the 'Each mass move'. if both masses are red then the tension can be artificially set from 0 to the minimum friction value between both masses (not the maximum value because this will cause the mass of minimum value to move and reduce the tension). If one mass is red and the other is green then the tension is the value of the green mass friction (minimum friction force).

First determine the possible motion direction according to creteria (a) and (b)

m sinβ = 20 sin53 = 16

M sinα = 40 sin30 = 20

The direction of the acceleration will be to the left.

Now check creterion (c) and (d) to verify is motion occures:

Motion equations for the case with acceleration to the left are:

On mass M:	ΣF_x = M g sinα − T − M g cosα μ_1k = M a	(3)
On mass m:	ΣF_x = T − m g sinβ − m g cosβ μ_2k = m a	(4)

Because we already found that the motion is to the left we have to check condition (d) only and we get 14.8 > 12.9 so the masses will slip to the left and the acceleration is:

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Find the acceleration of the system of masses neglecting the mass of the string and the inertia of the pulley.

From the forces diagram on mass m and M and assuming that the acceleration is to the right hance negative we get:

On mass m₁	ΣF_x = T₁ − m₁gμ₁ = m₁a	(1)
On mass m₂	ΣF_xy = T₂ + m₂g sin θ − T₁ − m₂ g μ₂ cosθ = m₂a	(2)
On mass m₃	ΣF_y = m₃ g − T₂ = m₃a	(3)

We get 3 equations with three unkowns T₁ T₂ and the acceleration a, solving for a we get:

System is at rest when a = 0:

m₃g − m₁gμ₁ − m₂gμ₂ cosα = 0

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Find the acceleration and the tension in the rope of the system of masses shown (M > m), neglecting the mass of the string and the inertia of the pulley assume that the static and kinetic friction coefficients are equal to μ₁ μ₂ and the slope angle θ are known.

Notice that the two masses are moving in the oposite directions so the friction forces are:.

Friction forces on mass M are:	f₁ = (M + m)gμ₁ cosθ
Friction forces on mass m are:	f₂ = mgμ₂ cosθ

From the forces diagram on mass m and M and assuming that the acceleration of mass M is downward and is equal to the acceleration of mass m in the upward direction we get:

On mass m	ΣF_x = T − f₂ − mg sinθ = ma	(1)
On mass M	ΣF_x = Mg sinθ − f₁ − f₂ − T = Ma	(2)

Subsitute the values of f₁ and f₂ into eq. 1 and 2 and eliminating T from both equations we get a:

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Find the acceleration of the system of masses neglecting the mass of the rope and the inertia of the pulley.

From the forces diagram on mass m and M and assuming that the acceleration is to the downward direction of mass M we get:

On mass M:	ΣF_y = Mg − T = Ma	(1)
On mass m:	ΣF_y = T − mg = ma	(2)

Solving for the acceleration we get:
And the tensions in the cables are:

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Find the acceleration of the system of masses neglecting the mass of the rope and the inertia of the pulley, the friction coefficient between mass M and the surface is μ.

From the forces diagram on mass m and M and assuming that the acceleration is to the downward direction and notice that the acceleration of mass M is twice the accelaretion of mass m because when mass M is moving a length of x mass m will travel only half this distance.

Forces on mass M:	ΣF_x = T − Mgμ = Ma₁	(1)
Forces on mass m:	ΣF_y = mg − T₁ = ma₂	(2)
From acceleration:	a₁ = 2a₂	(3)
From the pulley:	T₁ = 2T	(4)

We got four equations with four unkowns T T₁ a₁ and a₂ , solving the equations we get:

The accelerations are:
The tensions are:

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Find the acceleration of the system of masses neglecting the mass of the rope and the inertia of the pulleys.

Forces on pulley 1:	ΣF_y = 2T − Mg = Ma₁	(1)
Forces on mass m:	ΣF_y = mg − T = ma₂	(2)
Accelerations balance:	a₂ = 2a₁	(3)

The relation between the accelerations can be found by measuring the distances that the masses travels, while mass m is moving a distance of x mass M will move a distance of x/2 so the acceleration of M is half the acceleration of m.

Solving for a₁ a₂ and T we get:

And the tension in the rope is:

The tension T is uniform along the cable.

The value of T₁ is:

T₁ = 2T

Notice that if we apply a force F downward at mass m we could lift a mass of 2F, so this system multiplies the force by 2.

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Find the acceleration of the system of masses neglecting the mass of the string and the inertia of the pulley.

On mass m:	ΣF_x = T cosθ − (mg − T sinθ)μ = ma cosθ	(1)
On mass M:	ΣF_y = Mg − T = Ma	(2)

The term a cosθ in equation (1) is because the acceleration of the two masses are different, the acceleration of mass m is coincide with the direction of the tension T while the actual acceleration is along the x axis.

Acceleration of M is:
Acceleration of m is:	a_m = a cosθ (in the x direction)

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m₁		kg
m₂		kg
m₃		deg
a₁		m/s²
a₂		m/s²
a₃		m/s²
T₁		N
T₂		N

Acceleration directions
m₁	m₂	m₃

Find the acceleration of the system of masses neglecting the mass of the string and the mass of the pulley.

We assume arbitrary that all three accelerations are downward and are equal to a₁ a₂ and a₃ . Now we can write the forces acting on each mass as:

ΣF_y on mass m₁	m₁ g − T₁ = m₁ a₁	(1)
ΣF_y on mass m₂	m₂ g − T₂ = m₂ a₂	(2)
ΣF_y on mass m₃:	m₃ g − T₂ = m₃ a₃	(3)
ΣF_y on pulley B:	T₁ = 2 T₂	(4)

We can see that we have 5 unknowns and only 4 equations so we have to derive another equation from the accelerations, a₁ is equal to the acceleration of pulley B which sould be half the accelerations of a₂ and a₃ but in negative sign because when a₁ is going down pulley B is going up, but we chose the accelerations in the downward direction.

Σ Accelerations:	a₁ = −(a₂ + a₃) / 2	(5)
	a₃ = − 2a₁ − a₂

The minus sign is because a₁ is in the opposite direction to a₂ and a₃. Substitute the value of T₁ from eq. (4) and a₃ from eq. (5) to eq. (1) (2) and (3) to get the matrix form:

Solving by Cramer's rule we have:

D = 2 (m₂m₃ + m₂m₃) + m₁(m₃ + m₂) = m₁m₂ + m₁m₃ + 4m₂m₃

After division of the numerator and the denominator by m₁m₂m₃ we got another form for T₂

If all masses are the same and equal to M then:


	T₂ = 2 g M / 3

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from the forces diagram we can write the equilibrium equations:

On mass m₁	ΣF_x = T₁ = f₁ = m₁ g μ₁	(1)
On mass m₂	ΣF_x = T₂ − f₁ − f₂ = m₂ a
	T₂ − m₁ g μ₁ − (m₁ + m₂) μ₂ g = m₂ a	(2)
On mass m₃	ΣF_y = m₃ g − T₂ = m₃ a	(3)
		(4)

From eq. (3) T₂ can be found

T₂ = m₃(g − a) =

Q:	Given three masses m₁ equal to m, mass m₂ equal to 3m and m₃ equal to 2m, μ₁ = 0 and masses m₂ and m₃ are moving at a constant velocity. Find the value of μ₂, T₁ and T₂
S:	For steady velocity we need that a = 0 From the acceleration equation (4) we have: m₃ − m₁ μ₂ − m₂ μ₂ = 0 Because there is no friction between surfaces of masses m₁ and m₂ then: T₁ = 0 Because the acceleration is 0 the tension T₂ in cable m₃ is: T₂ = m₃ g

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Mass M is equal to 5 kg and mass m to 3 kg the dynamic friction coefficient between the surfaces is 0.12, at time t = 0 mass M has initial velocity of 5 m/s² find the acceleration of the system until stoppage and the distance traveled in the first two seconds.

According to the second law of motion F = m a, the total force to drive the system have to overcome the weight of mass m and the friction caused by moving mass M.

m g + μ M g = −(M + m)a	(1)
	(2)

Because m g > M g μ then the system will finally stop moving to the left and mass m will start to fall so v_t = 0

From acceleration equations we have:	v_t² = v₀² + 2aS
And the distance until stoppage (v_t = 0):

The time traveled as a function of acceleration and v₀ and remembering that v_t = 0 can be calculated from:

v_t = v₀ + a t

(3)

Substitute the value of a we get the total travel time until stoppage:

The acceleration until stoppage according to eq. (2) is:

S_(2s) = v₀ t + a t² / 2 = 5 * 2 − 4.4 * 4 / 2 = 1.2 m

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A force F is applied to mass m₂ which is sliding on a frictionless wheels. Mass m₁ is located on mass m₂ with a friction coefficient of μ₁ this nass is connected to mass m₃ by a frictionless pulley, the friction coefficient between surfaces m₂ and m₃ is μ₂. Find the minimum force F that will keep mass m₁ in place and the acceleration of the masses.

Friction force f₂ is effective only when the system is accelerates to the right:

f₂ = m₃ a μ₂

(1)

In order to keep the system of masses m₁ and m₃ at rest we need that:

m₁ a = m₃ g

(2)

(in this case f₁ = f₂ = 0 because there is no movement relative to mass m₂)

According to second law of motion:

F = (m₁ + m₂ + m₃) a

(3)

From eq. (2)
Substitute a into eq. (3)

Question:	Given three masses m₁ = 10 kg, m₂ = 2 kg and m₃ = 1 kg. Find the minimum force F that will keep masses m₁ and m₂ in place and the acceleration of the masses if the force F is twice as the force found.
Solution:	For steady velocity we need that a = 0 From the acceleration equation we have: m₃ − m₂ μ₂ + m₁ μ₂ = 0 Because there is no friction between surfaces of masses m₁ and m₂ then: T₁ = 0 Because the acceleration is 0 the tension T₂ in cable m₃ is: T₂ = m₃ g

▲

Find the acceleration of the system of masses neglecting the mass of the string and the inertia of the pulley.

The friction force value is:

f = μ m₁ g cos θ

When mass m₂ is moving a distance of x meters the mass m₁ will travel half that distance hence the acceleration will be half the acceleration of m₂ eq. (3).

From equilibrium of forces we have:

On mass m₁:	ΣF_x = T₁ − m₁ g sinθ − μm₁ g cos θ = m₁ a₁	(1)
On mass m₂:	ΣF_y = m₂ g − T₂ = m₂ a₂	(2)
Accelerations:	a₂ = 2a₁	(3)
From tensions:	T₂ = 2T₁	(4)

▲

Three masses are hanging on frictionless pulleys as shown in the figure. Find the acceleration of the masses.

We choose arbitrary the direction of the accelerations of masses m₁ and mass m₂ as upward and the acceleration of m₂ downward.

From the forces diagram on the masses we have.

On mass m₁	ΣF_y = T − m₁ g = m₁ a₁	(1)
On mass m₂:	ΣF_y = m₂ g − 2T = m₂ a₂	(2)
On mass m₃:	ΣF_y = T − m₃ g = m₃ a₃	(3)
From accelerations:		(4)

Note: when mass m₁ is moving x₁ distance upward and mass m₃ a distance of x₃ upward, the contribution to the displacement of mass m₂ is the sum of half of this displacements x₂ = (x₁ + x₃) / 2 in the downward direction.

The equations with the unknowns T, a₁, a₂ and a₃ can be solved by Cramer's rule or by direct substitution.

Write the equations in the matrix form

D = − m₂m₃ − 4m₁m₃ − m₁m₂ = − (m₁m₂ + m₂m₃ + 4m₁m₃)

Find the acceletations and the tension in the rope if all the masses are equal to m₁ = m₂ = m₃ = m

Notice that if m₁ = m₃ = m and m₂ = 2m then the accelerations will be 0 and T = m

If the result of the acceleration is negative then the motion is opposite to that chosen.

▲

Mass m₁ is located on a slope with an angle θ and friction coefficient of μ, the mass ia connected to a mass m₂ and to the wall with two pullies as shown. Find thq biggest value of m₂ that the system will be at rest, and the accelerations of the bodies.

The friction force value is:

f = μ m₁ g cos θ

When mass m₂ is moving a distance of x meters the mass m₁ will travel half that distance hence the acceleration will be half the acceleration of m₂ eq. (3).

From equilibrium of forces we have:

On mass m₁:	ΣF_x = T₁ − m₁ g sinθ − μm₁ g cos θ = m₁ a₁	(1)
On mass m₂:	ΣF_y = m₂ g − T₂ = m₂ a₂	(2)
Accelerations:	a₂ = 2a₁	(3)
From tensions:	T₁ = 2T₂	(4)

Substituting eq. (3) and (4) into eq. (1) and (2) and solving for a₁ and T₂ we get:

▲

Two pulleys A and B are connected as shown in the figure, the cable which holds pulleys A and B is connected to mass m₁ with an angle θ and is resting on a surface with friction coefficient of μ, pulley B is connected to mass m₂. Find the minimum weight of mass m₂ that will cause mass m₁ to move and the tension of the cable in this case, also find the equivalent and the direction of the force acting on pulley A.

The friction force acting on mass m₁ is:

f = (m₁ g − T₁ sin θ) μ

From the forces diagram on mass m₁ and m₂ and assuming that the acceleration is to the downward direction on mass m₂ we have:

On mass m₁	ΣF_x = T₁ cosθ − (m₁ g − T₁ sin θ) μ = m₁ a₁	(1)
On mass m₂:	ΣF_y = m₂ g − T₂ = m₂ a₂	(2)
From pulley B:	T₂ = 2T₁	(3)

Because at rest the system acceleration equals 0 then a₁ = a₂ = 0

From eq. (1)	T₁ cosθ − (m₁ g − T₁ sin θ) μ = 0	(1a)
From eq. (2) and (3)	T₂ = m₂ g → T₁ = m₂ g / 2	(1b)

Substitute eq. (1b) into (1a) to get the value of mass m₂

Substitute eq. (1b) into (1a) to get the value of mass m₂
And the tension T₁ in the cable is:

In order to find the value and the direction of the force acting on pulley A according to the force diagram

▲

Three masses are connected with a rope of 0 mass as shown in the figure, a force F is apply to mass m₃ , friction coefficient μ if present between mass m₁ and the surface that is tilted by θ deg. Find the value of the force F that will keep the system at rest, if the force F is cancelled find the acceleration of the system.

In order to determine the direction of the motion we have to analyze the relationship of the equations:

m₃ > m₂ + m₁ sinθ	Motion is m₃ down	(a)
m₃ < m₂ + m₁ sinθ	Motion is m₃ up	(b)
m₃ = m₂ + m₁ sinθ	No motion	(c)

Assume that the acceleration of mass m₃ is downward then:

On mass m₁	ΣF_x = T₁ − μ m₁ g cos θ − m₁ g sinθ = m₁ a	(1)
On mass m₂	ΣF_y = T₂ − m₂ g − T₁ = m₂ a	(2)
On mass m₃	ΣF_y = m₃ g + F − T₂ = m₃ a	(3)

The same equations when the acceleration is to the left are:

On mass m₁	ΣF_x = m₁ g sinθ − T₁ − μ m₁ g cosθ = m₁ a	(4)
On mass m₂	ΣF_y = m₂ g + T₁ − T₂ = m₂ a	(5)
On mass m₃	ΣF_y = T₂ − m₃ g − F = m₃ a	(6)

We got three equations with three unknowns T₁ T₂ and a.

Solving eq. (1) (2) and (3) according to Cramer's rule we get:

Solving eq. (4) (5) and (6) according to Cramer's rule we get:

The minimum force F nedded to apply when a = 0, when the motion is intended to be to the left is:

F = g ( m₁ sinθ + m₂ − m₃ ) − m₁μ_s g cosθ

The maximum force F nedded to apply when a = 0, when the motion is intended to be to the right is:

F = g ( m₁ sinθ + m₂ − m₃ ) + m₁μ_s g cosθ

In the case when F = 0 mass m₃ that keeps the system at rest is in the range of:

m₁ ( sinθ − μ cosθ ) + m₂ < m₃ < m₁ ( sinθ + μ cosθ ) + m₂

If all three masses are of the same weight M then the force F at rest should be in the range:

M g (sinθ − μ_s cosθ) < F < M g (sinθ + μ_s cosθ)

▲

Two masses m₁ and m₂ are connected throught 2 pulleys as shown in the figure the friction of m₁ and the surface is μ. Find the maximum and minimum values of m₂ that will steel keep the system at rest and the acceleration of m₂ when m₂ = 3m₁ the friction coefficient is μ = 0.1.

we assume that the acceleration of mass m₁ is upward along the slope, then the friction force direction will be downward as shown in the force diagram at right.

The friction force is equal to:

f = m₁ g cosθ μ

From the forces diagram on masses m and M we get the forces equations:

On mass m₁	ΣF_x = T₁ − m₁ g sinθ − m₁ g cosθ μ = m₁ a₁	(1)
On mass m₂:	ΣF_y = m₂ g − T₂ = m₂ a₂	(2)
From the pulley:	T₂ = 2T₁	(3)
From accelerations:	a₁ = 2a₂	(4)

Solving for T₁ we get:
And the acceleration is:		(need that a₂ > 0)
Mass m₂ is:
At rest a = 0 and mass m₂ is:	m₂ = 2 m₁ ( sinθ + μ cosθ )	(maximum value)

Note:

If the masses are moving at a constant speed then the acceleration is 0 and according to the equation v_t = v₀ + a t we can see that the bodies will keep moving at the speed of the initial velocity v_t = v₀.

If m₂ is bellow the minimum value then mass m₁ will slip down the slope and the friction force f will be upward as shown at left.

On mass m₁	ΣF_x = m₁ g sinθ − T₁ − m₁ g cosθ μ = m₁ a₁	(5)
On mass m₂:	ΣF_y = T₂ − m₂ g = m₂ a₂	(6)
From the pulley:	T₂ = 2T₁	(3)
From accelerations:	a₁ = 2a₂	(4)

Solving for T₁ we get:
And the acceleration is:		(need that a₂ > 0)
And the mass m₂ is:
At rest a = 0 and mass m₂ is:	m₂ = 2 m₁ ( sinθ − μ cosθ )	(minimum value)

Because there is friction in the system in order to find the acceleration of mass m₂ when m₂ = 3m₁ we have to determine the direction of the movment according to the following conditions.

(mass m₂ up) 2m₁(sinθ − μ cosθ) > m₂ > 2m₁(sinθ + μ cosθ) (mass m₂ down)

▲

Masses m₁ and m₂ are connected with two pullys as shown in the left figure, the friction coefficient of mass m₁ and the surface is μ find the acceleration of the masses and the tension in the cable connected to mass m₁.

The acceleration direction of mass m₂ will be downward.

From the forces diagram we have.

On mass m₁	ΣF_x = T₁ − m₁ g μ = m₁ a₁	(1)
On pulley A:	ΣF_y = 2 T₁ = T₂	(2)
	a₁ = 2 a₂	(3)
On pulley B:	ΣF_y = m₂ g − 2 T₂ = m₂ a₃	(4)
	a₂ = 2 a₃	(5)

We get 5 simple equations with 5 unknowns T₁ T₂ a₁ a₂ and a₃.

Solving for a₃ we get:
a₁ can be found by eq. (3) and (5)	a₁ = 2a₂ = 4a₃
Condition for movment (a₃ > 0):	m₂ > 4m₁ μ
And the tension T₁ from eq. (1) is:

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Mass m₁ is located on a slope with an angle of θ degree and a friction coefficient of μ, mass m₂ ia accelerating to the left direction on a frictionless wheels. Find the value of the acceleration that will keep mass m₁ at rest.

Because the force acting on mass m₂ is to the left according to Newtons law of motion mass m₁ produce a force opposite to the force on m₂.

The friction force value is:

f = N μ = (m₁ g cosθ + m₁ a sinθ) μ

From equilibrium of forces in the slope direction we have:

ΣF_x = m₁ g sinθ + (m₁ g cosθ + m₁ a sinθ) μ = m₁ a cosθ

And the acceleration is:

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A force F cause masses m₁ and m₂ to accelerate. mass m₂ is attached to mass m₁ only by friction force. Find the acceleration needed to prevent mass m₂ to slip down.

From the forces diagram on mass m₂ the friction force should be equal or more then the weight of mass m₂ in order not to slip, the friction force is a function of the acceleration and is equal to:

f = m₂ a μ_s

ΣF_y = m₂ a μ_s ≥ m₂ g

The forces acting on mass m₂ are:

ΣF_x = F − (m₁ + m₂)gμ₂ = (m₁ + m₂)a

Eliminating the value of F and substitute a = gμ₁ we get:

▲

Mass m is hanging on pulley A and is connected to two other pullys as shown in the figure, find a) the force F needed to keep mass m at rest b) the acceleration of the force downward and c) suppose that mass m ia the mass of a person lifting himselef what should be the force F that will keep him in place.

We choose the acceleration direction to cope with the direction of force F.

a) From the free body diagram of the pulleys we have.

On pulley A	ΣF_y = 2 T₁ = m g	(1)
On pulley B:	ΣF_y = 2 F = T₁	(2)
On pulley C:	ΣF_y = 2 F = T₂	(3)

From eq. (1) and (2) we get that: F = m g / 4 or m g = 4 F and T₁ = T₂

b) In order to find the acceleration of force F relative to the acceleration of mass m we have to analyze the distances that each pulley travels.

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A weight of 70 kgf is connected with three cables to a system of pulleys as shown in the left figure. Find the tension in each cable holding the weight.

We choose the tension of the right cable to be T and the forces diagram is drawn in the figure at right

The weigh is supported by three cables with a total tension of 2T + T / 2 + T = 3.5 T.

From equilibrium of the weight we have:

W = 3.5 T

T = W / 3.5 = 70 / 3.5 = 20N

And the tension in the cables are:

Verify the minimum force F needed to lift a weight of 100 kgf if the weight is connected by the series of pulleys as shown in the left figure.

At the right figure the tension in each cable started at the weight W is marked, it can easilly seen that each pulley is reducing the force F by half (n pulleys gives F = W / 2 ⁿ).

From equilibrium we have:

F = W / 16 = 100 / 16 = 6.25 kgf

▲

Three bodies are connected through a pulley as shown in the left figure, if the friction between the surface and mass m₁ is μ_s find the accelerations of the bodies.

We will mark the acceleration of mass m₂ and m₃ as a₂ this acceleration is the acceleration relative to the pulley. Therfore the real accelerations will be: a_m₂ = a₂ − a₁ and a_m₃ = a₁ + a₂

The kinetic friction force acting on mass m₁ is:

f = m₁ g μ

From the free body diagram on mass m₁ and assuming that the acceleration is to the right a₁:

On mass m₁	ΣF_x = T₁ − f = m₁ a₁	(1)
On mass m₂:	ΣF_y = T₂ − m₂ g = m₂ a_m₂	(2)
On mass m₃:	ΣF_y = m₃ g − T₂ = m₃ a_m₃	(3)
On the pulley:	T₁ = 2T₂	(4)

After substituting eq. (4) and the values for a_m₂ and a_m₃ we get the system of equations:

Solving this system of linear equations the determinant is equal to: D = 4 m₂ m₃ + m₁ ( m₃ + m₂ )

And by Cramer's rule we get:

▲

Three masses are connected to each other by ropes of 0 mass, a force F is apply to mass m₃. Find the value of the tension in the ropes and the acceleration of the masses if a friction coefficient of μ exists between the masses and the surface.

From the free body diagram the friction forces and the equilibrium equations of the masses are:

Friction forces:

f₁ = m₁ g μ

f₂ = m₂ g μ

f₃ = m₃ g μ

ΣF_x = T₁ − m₁ g μ = m₁ a	(1)
ΣF_x = T₂ − T₁ − m₂ g μ = m₂ a	(2)
ΣF_x = F − T₂ − m₃ g μ = m₃ a	(3)

We got three equations with three unknowns T₁ T₂ and a

According to Cramer's rule the value of the coefficients determinant is D = − (m₁ + m₂ + m₃)

Another way to solve the problem is to look on the three masses as one mass equal to M = m₁ + m₂ + m₃

From the free body diagram we have:	F − M g μ = M a	and a is:

Once we found the acceleration a we can use eq. (1) and (2) to find T₁ and T₂

▲

Two masses are connected by a rope of 0 mass to each other, the masses are placed on a slope of angle θ the friction coefficient between the masses and the surface are μ₁ and μ₂ respectively. Find a) the acceleration and the tension in the rope, b) the maximum angle that the masses will stay at rest.

Free body diagram:

Friction forces caused by the masses movment are:

f₁ = m₁ g μ₁

f₂ = m₂ g μ₂

From the free body diagram write the equilibrium equations of each mass:

ΣF_x = m₁ g sinθ − T − m₁ g cosθ μ₁ = m₁ a	(1)
ΣF_x = T + m₂ g sin θ − m₂ g cosθ μ₂ = m₂ a	(2)

We got two equations with two unknowns T and a. Solving by Cramer's rule we have:

D = −(m₁ + m₂)

The maximum angle for rest condition can be calculated by assuming that a = 0 and we get:

(3)

Note: In order to create tension in the rope we need that the lower mass acceleration will be bigger then the upper mass acceleration so we get:

m₁ g sinθ − m₁ g μ₁ cosθ = m₁ a_m₁	➞	a_m₁ = g(sinθ − μ₁ cosθ)
m₂ g sinθ − m₂ g μ₂ cosθ = m₂ a_m₂	➞	a_m₂ = g(sinθ − μ₂ cosθ)

From both equations it is clear that the mass with lower friction coefficient will have bigger acceleration and should be in the lower position.

Given two masses of 5 kg and 10 kg and a friction coefficients of 0.1 and 0.2 respectively. Find a) the order of the location of the masses that will cause a tension in the rope b) the minimum angle that will cause the masses to slipe and c) the acceleration of the masses if the angle is 30 degree.

a) From the privious note the lower mass should be the mass with lower friction coefficient m₁ = 5 kg

b) from eq (3) we can find the lowest angle that the masses will slipe (ΣF_x = 0).

c) if the slope angle is 30 degree then the ecceleration and the tension as found before will be:

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Mass m₁ is located on mass m₂ and has a friction coefficients of μ_1S and μ_1k the friction coefficients of m₂ and the lower surface are μ_2S and μ_2k. Find a) the biggest force F that can be applied to m₁ without causing mass m₁ to slip, b) if the applied force is twice the force found what are the accelerations of the masses.

There are 4 movements posibillitys for the masses:

1) No movement at all, this happens when:

F < m₁ g μ_1S

and

m₁ μ_1S < (m₁ + m₂) μ_2S (f₁ < f₂)

2) Only m₁ is moving, this happens when:

F > m₁ g μ_1S

and

m₁ μ_1S < (m₁ + m₂) μ_2S (f₁ < f₂)

The acceleration of m₁ is:

3) m₁ and m₂ are moving together, this happens when:

F > m₁ g μ_1S

and

m₁ μ_1S = (m₁ + m₂) μ_2S (f₁ = f₂)

The acceleration of m₁ is:

The maximum force F that the masses will move together is

4) m₁ and m₂ moves at different accelerations, this happens when:

F > m₁ g μ_1S

and

m₁ μ_1S > (m₁ + m₂) μ_2S (f₁ > f₂)

The acceleration of m₁ is:

The maximum friction forces acting on m₁ is

f₁ = m₁ g μ_1S

From the free body diagram mass m₁ will move when force F overcomes the static friction forc f₁ .

ΣF_x = F − m₁ g μ_1S = m₁ a₁

(1)

Once the force F is bigger then f₁ m₁ accelerates by:

ΣF_x = m₁ g μ₁ − (m₁ + m₂) g μ₂ = m₂a₂

(2)

We got two equations with two unknowns a₁ and a₂

Note: both accelerations must be positive should be in the direction of the force F so a₁ >= 0 and a₂ >= 0

a) In order to keep both masses moving together we have to fullfil the condition that a₁ = a₂

b) if the force is twice the force found in a) and is 2F then the accelerations of the bottom mass will be the same because the force acting on mass m₂ is only the friction force. The acceleration on mass m₁ will be:

Q.	A masse of 3 kg is mounted on top of a mass of 7 kg the friction coefficients between the masses are μ_1S = 0.8 and μ_1k = 0.1 and between the lower mass and the surface are μ_2S = 0.2 and μ_2k = 0.1. Find the maximum force F that will pul both masses together.

▲

The maximum friction forces acting on m₁ is

f₁ = m₁ g μ_1S

From the free body diagram mass m₁ will move when force F overcomes the static friction forc f₁ .

ΣF_x = F − m₁ g μ_1S = m₁ a₁

(1)

Once the force F is bigger then f₁ m₁ accelerates by:

ΣF_x = m₁ g μ₁ − (m₁ + m₂) g μ₂ = m₂a₂

(2)

We got two equations with two unknowns a₁ and a₂

Note: both accelerations must be positive should be in the direction of the force F so a₁ >= 0 and a₂ >= 0

a) In order to keep both masses moving together we have to fullfil the condition that a₁ = a₂

▲

Mass m₁ is moving at an initial speed of v₀ on mass m₂ the kinetic friction coefficient between mass m₁ and mass m₂ is μ₁ and between mass m₂ and the surface is μ₂. What is the distance travelled by mass m₁ relative to mass m₂ until mass m₁ stopes.

The friction force f₁ = m₁ g μ₁ is causing mass m₁ to decelerate from the initial velocity v₀ but this force is causing mass m₂ to accelerate after overcomming the friction force f₂ = (m₁ + m₂) g μ₂. The relative acceleration of mass m₁ relative to mass m₂ is a₁ − a₂

From the free body diagram the equilibrium equations are:

ΣF_x = m₁ g μ₁ = m₁ (a₁ − a₂)	(1)
ΣF_x = m₁ g μ₁ − (m₁ + m₂) g μ₂ = m₂a₂	(2)

We got two equations with two unknowns a₁ and a₂

From eq. (2)
From eq. (1)

The final velocity is v_t = 0 and we find the total travel distance by:

v_t² = v₀² − 2 a D

The minus sign is because deceleration is negative acceleration.

▲

Two masses m₁ and m₂ are located on the floor the masses then conected by a pulley as shown in the figure. Find the acceleration of the masses and the tension in the rope connecting the masses if a force of F is apply to the pulley upward, suppose that m₁ > m₂.

We mark the accelerations of the masses as a₁ and a₂, from the free body diagram we get:

On mass m₁:	ΣF_y = T − m₁ g = m₁ a₁	(1)
On mass m₂:	ΣF_y = T − m₂ g = m₂ a₂	(2)
On the pulley:	ΣF_y = F = 2 T	(3)

From eq. (3) we get:

T = F / 2

And substituting T to eq. (1) and (2) we get:

Notes: because accelerations a₁ and a₂ can not be negative (opposite to the direction shown) because of the floor there are some unique casses:

Samll mass m₂	F < 2 m₂ g	The force F is not enough to lift any one of the masses
Bigger mass m₁	2 m₂ g < F < 2 m₁ g	The force F will lift the small mass but not the bigger mass
	2 m₁ g < F	The force F will lift both masses

For example if mass m₁ = 30 kg and m₂ = 20 kg then find the accelerations when F = 100N, F = 400N and F = 1000N.

F [N]	a₁ [m/s²]	a₂ [m/s²]	Notes
100N	100 / 30 − 9.8 = − 6.5	100 / 20 − 9.8 = − 4.8	No lift of any mass a₁ = 0 a₂ = 0
200N	200 / 30 − 9.8 = − 3.1	200 / 20 − 9.8 = 0.2	Only small mass is lifted a₁ = 0
1000N	400 / 30 − 9.8 = 3.5	400 / 20 − 9.8 = 10.2	Both masses moves upward

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m₁		kg
m₂		kg
μ
F		N
θ		deg
a		m/s₂
T		N
f		N

Mass m₁ and m₂ are at rest when a force F is applied at an angle of θ degree. Find the maximum force F that the system will stay at rest and the acceleration of the system.

The friction force acting on m₁ is: f = N μ = (m₁ g + F sin θ) μ.

In order to determine the possible direction of the motion we have to check the forces acting on mass m₁ without friction force.

if F cosθ > m₂ g	m₁ moves to the left	(a)
if F cosθ < m₂ g	m₁ moves to the right	(b)

After we determined the direction of the motion we can set the right direction of the friction force and check if there is motion at all due to the friction force.

Let assume that the acceleration of m₁ is to the left, from the free body diagram we have

On m₁	F cosθ − f − T = m₁ a	(1)
On m₂	T − m₂ g = m₂ a	(2)

From eq. (1) and (2) we can find the values of a and T

If we asumme that the acceleration of mass m₁ is to the right direction the equilibrium equations will be as follows:

On m₁	T − F cosθ − f = m₁ a	(3)
On m₂	m₂ g − T = m₂ a	(4)

If we get a negative value for the acceleration in both directions then motion is not possible.

The conditions for motion are:

If	F cosθ − f > m₂ g	m₁ moves to the left	(c)
If	F cosθ + f < m₂ g	m₁ moves to the right	(d)

The range of the force F that the system is at rest is when it is equal to the friction force in both directions:

the static and dynamic friction coefficient between the surface and m₁ is μ = 0.4. Find the friction force acting on m₁ if the force equal 250N and θ = 30 degree and m₁ = 10kg and m₂ = 30 kg.

First we check conditions (c) and (d) to verify if the masses are moving.

Condition (c)	30 * g > 250 * cos30 + (10 * g + 250 * sin30) 0.4	False
Condition (d)	30 * g < 250 * cos30 - (10 * g + 250 * sin30) 0.4	False

Because both conditions are false the system is at rest.

a) Friction force can not be calculated by the equation for f because there is no movement, f will be calculated by the difference between force of mass m₂ g and F * cosθ

	*f = m₂ g − F cosθ** = 30 * 9.8 − 250 * cos30 = 77.5 N
b) maximum force F	F = g(100.4 + 30) /(cos 30 - sin 30 0.4) = 500.3 N
c) acceleration when F = 0	a = −g(10*0.4 - 30) /(10 + 30) = 6.4 m / s²

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Two masses m₁ and m₂ are located on a tilted slope of angle θ and with friction coefficient of μ. A force F is applied to the lower mass in the upward direction. Find the accelerations of the masses and the force that exerts by the masses on each other.

We will choose a positive acceleration upward.

The friction forces are:

f₁ = m₁ g μ cosθ

f₂ = m₂ g μ cosθ

From the forces diagram we can derive the equilibrium equations:

ΣF_x = F − P − f₁ − m₁ g sin θ = m₁ a	(1)
ΣF_y = P − f₂ − m₂ g sin θ = m₂ a	(2)

From eq. (1) and (2) we ca evaluate the values of a and P :

What will be the result of the acceleration and the force P if we change the locations of m₁ with m₂ and neglecting the friction assume that m₁ = m and m₂ = 3m and the force F = 4mg and the slope angle is 30 degree.

From the free body diagram the equilibrium equations of the bodies are:

ΣF_y = P − f₁ − m₁ g sin θ = m₁ a	(2)
ΣF_x = F − P − f₂ − m₂ g sin θ = m₂ a	(1)

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Two masses m₁ and m₂ are connected through 2 pulleys as shown in the figure the friction of m₁ and the surface is μ. Find the accelerations and the tensions in the ropes.

In order to find the direction of the motion we will analyze the forces on the pulley by neglecting the friction force, it is easy to see that if:

m₁ sinθ > 2 m₂	m₁ is moving downward	(a)
m₁ sinθ < 2 m₂	m₁ is moving upward	(b)
m₁ sinθ = 2 m₂	There will be no motion	(c)

If m₂ exceeds a maximum value then mass m₂ will start to move down and the friction force direction will be as shown in the free body diagram at right.

The friction force is equal to:

f = N₁ μ = m₁ g cosθ μ

From the forces diagram on mass m₁ and m₂ and assuming that the acceleration is to the downward direction of mass m₂ we have.

On mass m₁	ΣF_x = T₁ − f − m₁ g sinθ = m₁ a₁	(1)
On mass m₂:	ΣF_y = m₂ g − T₂ = m₂ a₂	(2)
From the accelerations:	a₂ = 2a₁	(3)
From the pulley tension:	T₁ = 2T₂	(4)